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# Solution: A sample of steam with a mass of 0.510 g and at a temperature of 100°C condenses into an insulated container holding 4.50 g of water at 2.0°C.( ΔH°vap = 40.7 kJ/mol, water = 4.18 J/g•°C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

###### Problem

A sample of steam with a mass of 0.510 g and at a temperature of 100°C condenses into an insulated container holding 4.50 g of water at 2.0°C.( ΔH°vap = 40.7 kJ/mol, water = 4.18 J/g•°C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

###### Solution

We’re being asked to determine the final temperature of the given mixture.

We will use the heat released by the sample of steam to calculate the final temperature of the mixture.

Recall that heat (q) can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{∆}}{\mathbf{T}}}$

q = heat, J

+qabsorbs heat
–qloses heat

m = mass (g)
c = specific heat capacity = J/(g·°C)
ΔT = Tf – Ti = (°C)

Based on the given system:

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